1+1+2+1+2+3+1+2+3+4+1+2+3+4+5+"""""""(省略号)1+2+3+""""""✀+n=?

2024-12-17 16:36:32
推荐回答(3个)
回答1:

=n*1+(n-1)*2+(n-2)*3+....+(n-(n-1))*n
=(1+2+3+...+n)*n-(1*2+2*3+3*4+...+(n-1)*n)
=n^2(n+1)/2-((1^2-1)+(2^2-2)+(3^2-3)+……+(n^2-n))
=n^2(n+1)/2-((1^2+2^2+3^2+……+n^2)-(1+2+3+……+n))
= n^2(n+1)/2-(n(n+1)(2n+1)/6-n(n+1)/2)
=n^2(n+1)/2-n(n+1)(2n+1)/6+n(n+1)/2
=(n+1)(3n^2-2n^2-n+3n)/6
=(n+1)(n^2+2n)/6
=n(n+1)(n+2)/6

回答2:

=n

回答3:

0